IPMAT Indore 2021Algebra > MediumEntered answer:✅ Correct Answer: 5310Related questions:If 112+122+132+…\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots121+221+321+… up to ∞=π26\infty = \frac{\pi^2}{6}∞=6π2, then the value of 112+132+152+…\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots121+321+521+… up to ∞\infty∞ isIf f(n)=1+2+3+⋯+(n+1)f(n)= 1 + 2 + 3 +\cdots+(n+1) f(n)=1+2+3+⋯+(n+1) and g(n)=∑k=1k=n1f(k)g(n)= \sum_{k=1}^{k=n} \dfrac{1}{f(k)}g(n)=∑k=1k=nf(k)1, then the least value of nnn for which g(n)g(n)g(n) exceeds the value 99100\dfrac{99}{100}10099 is:The sum of the first 15 terms in an arithmetic progression is 200, while the sum of the next 15 terms is 350. Then the common difference is